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(A) For a smooth inclined plane, the acceleration is $a_s = g \sin 	heta$. The time taken to cover distance $d$ is $t_s = \sqrt{2d / a_s} = \sqrt{2d

Vedclass · Accepted Answer

$3/4$

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(A) For a smooth inclined plane, the acceleration is $a_s = g \sin 	heta$. The time taken to cover distance $d$ is $t_s = \sqrt{2d / a_s} = \sqrt{2d / (g \sin 	heta)}$.For a rough inclined plane, the acceleration is $a_r = g(\sin 	heta - \mu \cos 	heta)$. The time taken to cover distance $d$ is $t_r = \sqrt{2d / a_r} = \sqrt{2d / (g(\sin 	heta - \mu \cos 	heta))}$.Given $t_r = t$ and $t_s = t/2$, we have $t_r = 2t_s$.Therefore, $\sqrt{2d / (g(\sin 	heta - \mu \cos 	heta))} = 2 \sqrt{2d / (g \sin 	heta)}$.Squaring both sides: $1 / (\sin 	heta - \mu \cos 	heta) = 4 / \sin 	heta$.$\sin 	heta = 4 \sin 	heta - 4 \mu \cos 	heta$.$4 \mu \cos 	heta = 3 \sin 	heta$.$\mu = (3/4) 	an 	heta$.Substituting $	heta = 45^{\circ}$, $\mu = (3/4) 	an 45^{\circ} = 3/4 	imes 1 = 3/4$.

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